KCC's quizzes AQQ235 about a bipolar common emitter amplifier - a kind proposal from our colleague Martin Walker

Thanks Giovanni for your feedback. You are right in saying that there is apparently no feedback, nor AC, nor DC. Usually one inserts a resistor in the emitter: this would insure a solid DC (negative) feedback in the circuit. But here...

The circuit is bad in the sense its behavior changes a lot with temperature. 

Having that said, the circuit works as amplifier. It does so thanks to the fact that the collector acts as a quite stable good current source; quite independent of supply change

yes, it acts as amplifier but the questions were:

1. Is there enough information to work out the gain of this circuit?NO, the gain is not fixed and depends by BJT parameters that have its own variation part to part and with temperature 
2. If so, what is the voltage gain of the circuit?Difficult to say with only the input data provided
3. Is it a good amplifier?NO, it is amplifier but not good. Temperature derivation can change the bias point so much that the output waveform can be distorced when the output signal is big enough. so surely it is NOT a good amplifier

Thanks

What you say is absolutely correct, in that the gain depends on the BJT parameters which vary greatly with temperature.  However, it's a bit of a tricksy question because the temperature is specified at 25degC.  We've also not explicitly specified the bias, other than it is implicit in Vout being at 1/2 Vcc or 3V under quiescent conditions.

The gain of a grounded common emitter amplifier is just -gmRc and gm=Ic/(kT⁄q).  Rc is just Vcc-Vout/Ic.  Vout-Vcc=3 and the Ic's cancel each other out, so we end up with small-signal gain as 3/kT/q.  kT/q is roughly 25mV at 25degC, so this works out to be a gain of about 120 or 41dB.

Of course, it's a terrible amplifier because the gain is both temperature and input signal dependent.  The input impedance and loading of the input signal also varies as this is affected by the re and beta of the BJT.  This is why most common emitter amplifiers have an emitter resistor to provide negative feedback to reduce and stabilise the gain.

As I say. it's a bit of a tricksy question but I'm glad everyone had some fun with it.  I first came across this brain-teaser in a video by Paul Brokaw on ADITube.  The link is further up the comments and is well worth a look.

1. There is sufficient information regarding the gain of the circuit. 

2. Voltage gain Av = (-3B)/(26mV) where B is the Beta of the circuit. 

3. Seems like the amplifier is highly dependent on Vbe of the circuit, which limits the design on the specified temperature ( which is where I got the 26mV ) - a parameter that varies as soon as power is drawn from the supply. 

4. One thing to add is to simply put an emitter resistor to act as a voltage feedback in its input, thus reducing the design's dependency on Vbe. 

Woops, something went wrong with my model, forgot that the input impedance is Bre and not just re. So B cancels out.

Great Keenan!

Gain=-3/25m

Correct. Close to -120 : congratulations Mpara